The sum to the $$n$$ term of the series $${\text{cose}}{{\text{c}}^{ - 1}}\sqrt {10} + {\text{cose}}{{\text{c}}^{ - 1}}\sqrt {50} + {\text{cose}}{{\text{c}}^{ - 1}}\sqrt {170} + ..... + {\text{cose}}{{\text{c}}^{ - 1}}\sqrt {\left( {{n^2} + 1} \right)\left( {{n^2} + 2n + 2} \right)} {\text{ is}}$$

Solution :
$$\eqalign{ & {\text{Let, }}\theta = {\text{cose}}{{\text{c}}^{ - 1}}\sqrt {\left( {{n^2} + 1} \right)\left( {{n^2} + 2n + 2} \right)} \cr & \Rightarrow {\text{cose}}{{\text{c}}^2}\theta = \left( {{n^2} + 1} \right)\left( {{n^2} + 2n + 2} \right) \cr & = {\left( {{n^2} + 1} \right)^2} + 2n\left( {{n^2} + 1} \right) + {n^2} + 1 \cr & = {\left( {{n^2} + n + 1} \right)^2} + 1 \cr & \Rightarrow {\cot ^2}\theta = {\left( {{n^2} + n + 1} \right)^2} \cr & \Rightarrow \tan \theta = \frac{1}{{{n^2} + n + 1}} = \frac{{\left( {n + 1} \right) - n}}{{1 + \left( {n + 1} \right)n}} \cr & \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\frac{{\left( {n + 1} \right) - n}}{{1 + \left( {n + 1} \right)n}}} \right] = {\tan ^{ - 1}}\left( {n + 1} \right) - {\tan ^{ - 1}}n \cr} $$
Thus, sum $$n$$ terms of the given series
$$\eqalign{ & = \left( {{{\tan }^{ - 1}}2 - {{\tan }^{ - 1}}1} \right) + \left( {{{\tan }^{ - 1}}3 - {{\tan }^{ - 1}}2} \right) + \left( {{{\tan }^{ - 1}}4 - {{\tan }^{ - 1}}3} \right) + ..... + \left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right) \cr & \Rightarrow {\tan ^{ - 1}}\left( {n + 1} \right) - \frac{\pi }{4} \cr} $$